On increasing the radii of the base and the height of a cone by 20%, its volume will increase by

(d) 72.8%
Let the original radius of the cone be r and height be h.

Then, original volume $=\frac{1}{3} \pi r^{2} h$

Let $\frac{1}{3} \pi r^{2} h=V$

New radius $=120 \%$ of $r$

$=\frac{120 r}{100}$

$=\frac{6 r}{5}$

New height = 120% of h

$=\frac{120 h}{100}$

$=\frac{6 h}{5}$

Hence, the new volume $=\frac{1}{3} \pi \times\left(\frac{6 \mathrm{r}}{5}\right)^{2} \times \frac{6 \mathrm{~h}}{5}$

$=\frac{216}{125}\left(\frac{1}{3} \pi r^{2} h\right)$

$=\frac{216}{125} \mathrm{~V}$

Increase in volume $=\left(\frac{216}{125} \mathrm{~V}-\mathrm{V}\right)$

$=\frac{91 \mathrm{~V}}{125}$

Increase in $\%$ of the volume $=\left(\frac{91 \mathrm{~V}}{125} \times \frac{1}{\mathrm{~V}} \times 100\right) \%$

$=72.8 \%$